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Chapter 28: Problem 56

An electron is moving in the vicinity of a long, straight wire that lies along the \(x\) -axis. The wire has a constant current of 9.00 A in the \(-x\) -direction. At an instant when the electron is at point \((0,0.200 \mathrm{~m}, 0)\) and the electron's velocity is \(\overrightarrow{\boldsymbol{v}}=\left(5.00 \times 10^{4} \mathrm{~m} / \mathrm{s}\right) \hat{\imath}-\left(3.00 \times 10^{4} \mathrm{~m} / \mathrm{s}\right) \hat{\jmath},\) what is the force that the wire exerts on the electron? Express the force in terms of unit vectors, and calculate its magnitude.

Short Answer

Expert verified
The force that the wire exerts on the electron is calculated in terms of unit vectors first, and then its magnitude is obtained by applying the Pythagorean theorem. It is important to note that the direction of the magnetic field and force is determined using the right-hand rule.

Step by step solution

01

Calculate the Magnetic Field Due to the Wire

The wire generates a magnetic field that is circular and whose intensity can be computed using Ampere's circuital law: \[ B = 渭_0 * I / 2蟺r \] where \( 渭_0 \) is the permeability of free space and has a value of \( 4蟺 * 10^{-7} \, Tm/A \), \( r \) is the distance between the wire and the point where the field is being computed (0.2 m in this case), and \( I \) is the current (9 A). The direction of the magnetic field at the location of the electron can be determined using the right-hand rule and is \(\hat{k}\) (outward from the plane).
02

Compute the Magnetic Force

The magnetic force on the electron can be computed using the formula \[ F = q(v 脳 B) \] This is a vector product, so the direction of the force is perpendicular to both the velocity of the electron and the magnetic field. The charge of an electron is \(-1.6 * 10^{-19} \, C\). Let's perform the multiplication \(v 脳 B\) to get the force vector.
03

Compute the Magnitude of the Force

The magnitude of the force can be obtained from the force vector simply by computing the vector's length using the Pythagorean theorem if the force vector's components along \(\hat{\imath}\) and \(\hat{\jmath}\) are \(F_x\) and \(F_y\) respectively. The magnitude is then \[ F = \sqrt{F_x^2 + F_y^2} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetic Field
When a current flows through a wire, it generates a magnetic field around it. This field forms concentric circles around the wire. The strength of this magnetic field depends on several factors, such as the amount of current flowing through the wire and the distance from the wire. Understanding the concept:
  • The magnetic field created by a straight wire can be determined using the formula: \[ B = \frac{渭_0 \, I}{2蟺r} \] where \( B \) is the magnetic field, \( 渭_0 \) is the permeability of free space \( (4蟺 \times 10^{-7} \, Tm/A) \), \( I \) is the current, and \( r \) is the radial distance from the wire.
  • In this exercise, the wire has a current of 9 A. The electron is located 0.2 m from the wire. Plug these values into the formula to calculate the magnetic field.
  • The direction of the magnetic field can be determined using the right-hand rule: For the given scenario, the magnetic field emerges out of the plane, pointing towards \( \hat{k} \).
Ampere's Circuital Law
Ampere's Circuital Law helps us to calculate the magnetic field generated by electric currents. In simple terms, this law states that the total magnetic field around a closed path is proportional to the current flowing through that path.Key points about Ampere's Circuital Law:
  • Ampere's Circuital Law is mathematically represented as: \[ \oint \overrightarrow{B} \cdot d\overrightarrow{l} = 渭_0 \, I_{enc} \] where the left side of the equation is a line integral around a closed loop, and \( I_{enc} \) is the total current enclosed by the path.
  • This principle is vital for calculating the magnetic field in cases with high symmetry, such as long straight wires, solenoids, and toroids.
  • In practical problems involving finite straight wires, we often use a simplified version, relying on the derived formula for magnetic field around an infinite wire.
Vector Product
The vector product, often referred to as the cross product, is a calculation that helps us find a vector that is perpendicular to two given vectors. For this exercise, the force on the electron is determined using a vector product.Explore the cross product:
  • The formula used is: \[ F = q(\overrightarrow{v} \times \overrightarrow{B}) \] where \( q \) is the charge of the electron, \( \overrightarrow{v} \) is the velocity vector, and \( \overrightarrow{B} \) is the magnetic field vector.
  • The cross product calculates a vector that is perpendicular to both the voltage and the magnetic field.
  • Use determinant methods or unit vectors to solve the cross product, giving directional components of the resultant force vector.
The cross product's result will allow you to find the direction and magnitude of the magnetic force acting on the electron.
Electron Velocity
Understanding the velocity of an electron in a magnetic field context is crucial for calculating the force it experiences. Velocity is a vector quantity, meaning it has both magnitude and direction.Key components of electron velocity in this context:
  • In this exercise, the electron's velocity is expressed in unit vector notation: \( \overrightarrow{v} = (5.00 \times 10^4 \, m/s)\hat{\imath} - (3.00 \times 10^4 \, m/s)\hat{\jmath} \). This indicates the electron is moving largely in the x-direction with a smaller component in the negative y-direction.
  • When calculating the magnetic force on the electron, this velocity vector will be used in the cross product formula \( \overrightarrow{v} \times \overrightarrow{B} \) to determine how the electron moves in relation to the magnetic field.
  • The speed and direction impact the force's direction, making it crucial to consider both elements when performing calculations.

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Most popular questions from this chapter

A long solenoid with 60 turns of wire per centimeter carries a current of 0.15 A. The wire that makes up the solenoid is wrapped around a solid core of silicon steel \(\left(K_{\mathrm{m}}=5200\right) .\) (The wire of the solenoid is jacketed with an insulator so that none of the current flows into the core.) (a) For a point inside the core, find the magnitudes of (i) the magnetic field \(\overrightarrow{\boldsymbol{B}}_{0}\) due to the solenoid current; (ii) the magnetization \(\overrightarrow{\boldsymbol{M}}\) (iii) the total magnetic field \(\overrightarrow{\boldsymbol{B}}\). (b) In a sketch of the solenoid and core, show the directions of the vectors \(\overrightarrow{\boldsymbol{B}}, \overrightarrow{\boldsymbol{B}}_{0},\) and \(\overrightarrow{\boldsymbol{M}}\) inside the core.

To use a larger sample, the experimenters construct a solenoid that has the same length, type of wire, and loop spacing but twice the diameter of the original. How does the maximum possible magnetic torque on a bacterium in this new solenoid compare with the torque the bacterium would have experienced in the original solenoid? Assume that the currents in the solenoids are the same. The maximum torque in the new solenoid is (a) twice that in the original one; (b) half that in the original one; (c) the same as that in the original one; (d) one-quarter that in the original one.

A solid conductor with radius \(a\) is supported by insulating disks on the axis of a conducting tube with inner radius \(b\) and outer radius \(c\) (Fig. E28.39). The central conductor and tube carry equal currents \(I\) in opposite directions. The currents are distributed uniformly over the cross sections of each conductor. Derive an expression for the magnitude of the magnetic field (a) at points outside the central, solid conductor but inside the tube \((ac)\)

A long, straight wire with a circular cross section of radius \(R\) carries a current \(I\). Assume that the current density is not constant across the cross section of the wire, but rather varies as \(J=\alpha r,\) where \(\alpha\) is a constant. (a) \(\mathrm{By}\) the requirement that \(J\) integrated over the cross section of the wire gives the total current \(I,\) calculate the constant \(\alpha\) in terms of \(I\) and \(R .\) (b) Use Ampere's law to calculate the magnetic field \(B(r)\) for (i) \(r \leq R\) and (ii) \(r \geq R .\) Express your answers in terms of \(I\).

We can estimate the strength of the magnetic field of a refrigerator magnet in the following way: Imagine the magnet as a collection of current-loop magnetic dipoles. (a) Derive the force between two current loops with radius \(R\) and current \(I\) separated by distance \(d \ll R\). Very close to the wire its magnetic field is about the same as for an infinitely long wire, and Eq.( 28.11 ) can be used. (b) Using Eq. ( 28.17 ), express the current \(I\) in terms of the magnetic field at the middle of the loop, and express the radius \(R\) in terms of the area of the loop. In this way, derive an expression for the force \(F\) between two identical current loops separated by a small distance \(d\) in terms of their mutual area \(A\) and center magnetic field \(B\). (c) Rearrange your result to obtain an expression for the magnetic field of a dipole with area \(A\) in terms of the force \(F\) from an identical dipole separated by a small distance \(d\). (d) Now notice that the force it takes to separate one magnet from your refrigerator is nearly the same as the force it takes to separate two magnets stuck together. Estimate that force \(F\). (e) Estimate the area of a refrigerator magnet. (f) Assume that when these magnets are stuck together or to the refrigerator, they are separated by an effective distance \(d=25 \mu \mathrm{m}\). Use the formula derived above to estimate the magnetic field strength of the magnet.
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